Given a standardized normal distribution (with a mean of 0 and a standard deviat

Question

Given a standardized normal distribution (with a mean of 0 and a standard deviation of 1), complete parts (a) through (d) EB Click here to view page 1 of the cumulative standardized normal distribution table. ?Click here to view page 2 of the cumulative standardized normal distribution table a. What is the probability that Z is less than 1.09? The probabity that Z is less than 1.09 is (Round to four decimal places as needed.) b. What is the probability that Z is greater than -0.26 The probability that Z is greater than -0.26 is Round to four decimal places as needed.) c. What is the probability that Z is less than -0.26 or greater than the mean? The probability that Z is less than -0.26 or greater than the mean is (Round to four decimal places as needed.) d. What is the probability that Z is less than -0.26 or greater than 1.09? The probability that Z is less than -0.26 or greater than 1.09 is (Round to four decimal places as needed.) Enter your answer in each of the answer boxes.

Explanation / Answer

(a) To find P(Z < 1.09):
Given Table gives for Z = 1.09, area = 0.8671

So,

P(Z < 1.09) = 0.8671

(b)

To find P(Z > - 0.26)

Given Table gives for Z = - 0.26, area = 0.4083

So,

P(Z>-0.26) = 1 - 0.4083 = 0.5987

(c)

P(Z < - 0.26 OR Z > mean) = P(Z<-0.26) + P(Z > mean) = 0.4083 + 0.5 = 0.9083

(d)

P(Z>1.09) = 1 - P(Z<1.09) =1 - 0.8671 = 0.1329

P(Z < - 0.26 OR Z > 1.09) = P(Z < - 0.26) + P(Z> 1.09) = 0.4083 + 0.1329 = 0.5412

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